The included angle between every two time points on the clock is 30 degrees (the rounded corner is 360 degrees divided by 12), so at four o'clock, the included angle between the hour hand and the minute hand is 120 degrees (4 times 30). After a period of time, the two hands overlap, that is, the angle traveled by the minute hand minus the angle traveled by the hour hand is 120. Also, you should know that if the minute hand goes 6 degrees per minute and the hour hand goes 0.5 degrees per minute, then the equation can be listed: 6x-0.5x= 120, that is, 5.5x= 120, and the solution is x = 240/1,kloc-0/. I hope I can help you.

The second question is the same. At five o'clock, the angle between the two hands is 150 degrees (30 times 5), and 150 degrees is greater than 90 degrees. So the first right angle should be before the two hands meet (that is, the minute hand is on the right). Minute hand and hour hand minus 150 degrees, it is 90 degrees. Equation 6544 can be listed.

After writing, my hands are at right angles again. At this time, the minute hand should meet the hour hand once (that is, the minute hand is on the left). Then the data goes on. The minute hand minus 150 degrees, and the hour hand minus it should be 90 degrees. The equation 6x- 150-0.5x=90 can be listed, and the solution is x = 480/6540.

Oh, that's it. If you don't know much, you can try to draw your own pictures. If you let me upload photos ... I think it will be more difficult to understand.

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